Question

In the given circuit diagram, a wire is joining points $\text{B}$ and $\text{D}$. The current in this wire is:

• A. $0.4\text{A}$
• B. $2\text{A}$
• C. $4\text{A}$
• D. zero

Answer 1

The correct option is B $2\text{A}$

From circuit diagram,
the effective resistance of $1\Omega$ and $4\Omega$ is,
$\frac{1}{R_1}=\frac{1}{1}+\frac{1}{4}\Rightarrow R_1=\frac{4}{5}$

And, the effective resistance of $2\Omega$ and $3\Omega$ is,

$\frac{1}{R_2}=\frac{1}{2}+\frac{1}{3}\Rightarrow R_2=\frac{6}{5}$


$R_{\text{eff}}=R_1+R_2=\frac{4}{5}+\frac{6}{5}=2\Omega$

$i=\frac{V}{R_{\text{eff}}}=\frac{20}{2}=10\text{A}$

From current distribution diagram,

$\therefore I_{BC}=\frac{4i}{5}-\frac{3i}{5}=\frac{i}{5}=2\text{A}$

Hence, $\left(B\right)$ is the correct answer.